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3 years ago

❥SCIENCE SUBJECT

Biology

1 answer:

Fiesta28 [93]3 years ago

4 0

9514 1404 393

Answer:

time: 1.122 seconds
range: 10.693 m
maximum height: 1.543 m

Explanation:

<u>Given</u>:

runner is launched at 30° angle to horizontal at 11 m/s

acceleration due to gravity is g = -9.8 m/s²

<u>Find</u>:

runner's hang time

runner's distance to the landing point

runner's maximum height

<u>Solution</u>:

The (horizontal, vertical) speed components will be ...

(11 m/s)(cos(30°), sin(30°)) = (5.5√3 m/s, 5.5 m/s)

The time of flight can be found from the height formula:

h(t) = 1/2gt² +vt . . . . . . where v is the vertical speed at launch

The time we're concerned with is the time when h(t)=0 and t>0.

0 = -4.9t^2 +5.5√3t = t(-4.9t +5.5√3)

The second factor is zero when ...

t = (5.5√3)/4.9 ≈ 1.122 . . . seconds hang time

The distance to the landing point will be the product of horizontal speed and hang time:

d = (5.5 m/s)(5.5√3/4.9 s) ≈ 10.693 m . . . . distance to landing

The maximum height can be found from the formula (based on conversion of kinetic energy to potential energy) ...

h = v²/|2g| = (5.5 m/s)²/(2(9.8 m/s²)) ≈ 1.543 m . . . . maximum height

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Subject: Biology

Topic: The cell

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